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题目链接：http://poj.org/problem?id=3070
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
题意描述：菲波那契数列可以用题中那个矩阵计算得到，给出n（n可以超大），求出Fn。
算法分析：赤果果的矩阵快速幂。如果对矩阵快速幂不熟悉的话，可以先百度看看讲解。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<algorithm> 7 using namespace std; 8 const int mod=10000; 9 10 int n; 11 struct matrix 12 { 13 int an[2][2]; 14 }res; 15 matrix multiply(matrix a,matrix b) 16 { 17 matrix s; 18 memset(s.an,0,sizeof(s.an)); 19 for (int i=0 ;i<2 ;i++) 20 { 21 for (int j=0 ;j<2 ;j++) 22 { 23 for (int k=0 ;k<2 ;k++) 24 { 25 s.an[i][j] += a.an[i][k]*b.an[k][j]; 26 if (s.an[i][j]>mod) s.an[i][j]%=mod; 27 } 28 } 29 } 30 return s; 31 } 32 void calc(int n) 33 { 34 matrix sum; 35 for (int i=0 ;i<2 ;i++) 36 { 37 for (int j=0 ;j<2 ;j++) 38 sum.an[i][j]= i==j ? 1 : 0 ; 39 } 40 for (int i=0 ;i<2 ;i++) 41 { 42 for (int j=0 ;j<2 ;j++) 43 res.an[i][j]=1; 44 } 45 res.an[1][1]=0; 46 while (n) 47 { 48 if (n&1) sum=multiply(sum,res); 49 n >>= 1; 50 res=multiply(res,res); 51 } 52 printf("%d\n",sum.an[1][0]%mod); 53 } 54 55 int main() 56 { 57 while (scanf("%d",&n)!=EOF && n!=-1) 58 { 59 calc(n); 60 } 61 return 0; 62 }
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原文地址：http://www.cnblogs.com/huangxf/p/4400942.html
